7.3. We need to find a bijective function between the two sets. }\] The concept of cardinality can be generalized to infinite sets. Finite Sets: Consider a set $A$. Hence, the function $$f$$ is injective. Hence, there is a bijection between the two sets. * The set of algebraic numbers. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. The mapping between the two sets is defined by the function $$f:\left( {0,1} \right] \to \left( {0,1} \right)$$ that maps each term of the sequence to the next one: ${f\left( {{x_n}} \right) = {x_{n + 1}},\;\text{ or }\;}\kern0pt{\frac{1}{n} \to \frac{1}{{n + 1}}. 3. Since $$f$$ is both injective and surjective, it is bijective. We see that the function $$f$$ is surjective. {n + m = b} The Cartesian product of an infinite number of sets, each containing at least two elements, is either empty or infinite; if the axiom of choice holds, then it is infinite. A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number. Assume that $${x_1} \ne {x_2}$$ but $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right).$$ Then, \[{\frac{1}{\pi }\arctan {x_1} + \frac{1}{2} }={ \frac{1}{\pi }\arctan {x_2} + \frac{1}{2},}\;\; \Rightarrow {\frac{1}{\pi }\arctan {x_1} = \frac{1}{\pi }\arctan {x_2},}\;\; \Rightarrow {\arctan {x_1} = \arctan {x_2},}\;\; \Rightarrow {\tan \left( {\arctan {x_1}} \right) = \tan \left( {\arctan {x_2}} \right),}\;\; \Rightarrow {{x_1} = {x_2},}$. Cardinality of a set is a measure of the number of elements in the set. In ZF, a set is infinite if and only if the power set of its power set is a Dedekind-infinite set, having a proper subset equinumerous to itself. In this case, we write $$A \sim B.$$ More formally, $A \sim B \;\text{ iff }\; \left| A \right| = \left| B \right|.$, Equinumerosity is an equivalence relation on a family of sets. For us it will suffice to distinguish between two types of infinite sets: countably infinite sets and uncountable (or uncountably infinite) sets… There are many sets that are countably infinite, ℕ, ℤ, 2ℤ, 3ℤ, nℤ, and ℚ. }\], The preimage $$x$$ lies in the domain $$\left( {a,b} \right)$$ and, ${f\left( x \right) = f\left( {a + \frac{{b – a}}{{d – c}}\left( {y – c} \right)} \right) }={ c + \frac{{d – c}}{{b – a}}\left( {\cancel{a} + \frac{{b – a}}{{d – c}}\left( {y – c} \right) – \cancel{a}} \right) }={ c + \frac{\cancel{d – c}}{\cancel{b – a}} \cdot \frac{\cancel{b – a}}{\cancel{d – c}}\left( {y – c} \right) }={ \cancel{c} + y – \cancel{c} }={ y.}$. We first discuss cardinality for finite sets and then talk about infinite sets. Two infinite sets $$A$$ and $$B$$ have the same cardinality (that is, $$\left| A \right| = \left| B \right|$$) if there exists a bijection $$A \to B.$$ This bijection-based definition is also applicable to finite sets. This poses few difficulties with finite sets, but infinite sets require some care. If an infinite set is partitioned into finitely many subsets, then at least one of them must be infinite. It is interesting to compare the cardinalities of two infinite sets: $$\mathbb{N}$$ and $$\mathbb{R}.$$ It turns out that $$\left| \mathbb{N} \right| \ne \left| \mathbb{R} \right|.$$ This was proved by Georg Cantor in $$1891$$ who showed that there are infinite sets which do not have a bijective mapping to the set of natural numbers $$\mathbb{N}.$$ This proof is known as Cantor’s diagonal argument.

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